Quadratic Equations and Their Properties

Standard Form of a Quadratic Equation

• Definition: The standard form of a quadratic equation is given by:
  $f(x) = ax^2 + bx + c$

• Parabola Direction:

  • If a > 0, the parabola opens upward.

  • If a < 0, the parabola opens downward.

• Stretch/Compression:

  • The absolute value of $a$ affects the width of the parabola.

  • If a > 1, the parabola is narrower.

  • If |a| < 1, the parabola is wider.

• Y-Intercept:

  • The value of $c$ represents the y-intercept of the graph, where the graph intersects the y-axis.

• Vertex:

  • The x-coordinate of the vertex can be found using the formula:
    $x = -\frac{b}{2a}$

Vertex Form of a Quadratic Equation

• Definition: The vertex form of a quadratic equation is:
  $f(x) = a(x - h)^2 + k$

• Vertex:

  • The vertex of the parabola is located at the point $(h, k)$.

• Axis of Symmetry:

  • The axis of symmetry is the vertical line represented by the equation:
    $x = h$

• Parabola Direction:

  • Similar to the standard form, if a > 0, the parabola opens upward, and if a < 0, it opens downward.

Average Rate of Change from a Function

• Concept: The average rate of change of a function over an interval provides a measure of how much the function's value changes on that interval.

• Example Calculation: Find the average rate of change of the function $f(x) = x^2 + 4$ on the interval $[1, 5]$.

  1. Calculate the function values at the endpoints of the interval:

  • For $x = 5$:
    $f(5) = (5)^2 + 4 = 25 + 4 = 29$

  • For $x = 1$:
    $f(1) = (1)^2 + 4 = 1 + 4 = 5$

  1. Apply the average rate of change formula:

  • Average rate of change:
    $\frac{29 - 5}{5 - 1} = \frac{24}{4} = 6$

  1. Result: The average rate of change is 6.

Vertical Motion Word Problems

• Standard Formula for Vertical Motion: $h(t) = -16t^2 + V<em>0t + h</em>0$

  • Where:

  • $V_0$ = initial vertical velocity

  • $h_0$ = initial height

Example Problem 1: Football Kick

• Scenario: A football is kicked from the ground with an initial vertical velocity of 48 ft/s.

• Objective: Determine how long it will be before it hits the ground.

  • The height function becomes:
    $h(t) = -16t^2 + 48t + 0$

  • To find the time when it hits the ground, set $h(t) = 0$:
    $0 = -16t(t - 3)$

  • Solve for $t$:

  • $0 = -16t <br>ightarrow t = 0 ext{ sec}$

  • $0 = t - 3 <br>ightarrow t = 3 ext{ sec}$

  • Conclusion: The football will hit the ground after 3 seconds.

Example Problem 2: Shot Put Throw

• Scenario: An athlete throws a shot put with an initial vertical velocity of 38 ft/s from a height of 5 ft.

• Objective: How high above the ground is it after 2 seconds?

  • The height function:
    $h(t) = -16t^2 + 38t + 5$

  • Substitute $t = 2$ into the equation:
    $h(2) = -16(2)^2 + 38(2) + 5$
    $h(2) = -64 + 76 + 5 = 17 ext{ ft}$

  • Result: The shot put is 17 ft above the ground after 2 seconds.

Finding Maximum/Minimum Values of a Parabola

From Standard Form

1. Determine the direction of the parabola by examining the leading coefficient $a$ in the quadratic equation:

   • If a > 0, the parabola opens upward, and the vertex represents a minimum point.

   • If a < 0, the parabola opens downward, and the vertex represents a maximum point.

2. Find the vertex:

   • The x-coordinate of the vertex is given by the formula:
     $x = -\frac{b}{2a}$

3. Calculate the maximum or minimum value:

   • Substitute the x-value of the vertex back into the original function to find the corresponding y-coordinate.

   • This y-value is the minimum or maximum value of the function.

From Vertex Form

1. Identify the 'k' value:

   • The 'k' value is the constant term added or subtracted outside the parentheses in the vertex form equation.

   • Example: In $y = 2(x - 3)^2 + 5$, the 'k' value is 5.

2. Determine if it's a minimum or maximum:

   • Check the coefficient 'a' in front of the parentheses:

     • If a > 0, the parabola opens upwards, and 'k' is the minimum value.

     • If a < 0, the parabola opens downwards, and 'k' is the maximum value.